Question

2 In the region shown in Figure 13.64 there is an electric field due to charged objects not shown in the diagram. A tiny glass ball with a charge of$\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{C}$placed at location A experiences a force of$\mathbf{(}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{,}\mathbf{-}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{N}$, as shown in the figure. (a) Which arrow in Figure 13.65 best indicates the direction of the electric field at location A? (b) What is the electric field at location A? (c) What is the magnitude of this electric field? (d) Now the glass ball is moved very far away. A tiny plastic ball with charge$\mathbf{-}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{C}$is placed at location A. Which arrow in Figure 13.65 best indicates the direction of the electric force on the negatively charged plastic ball? (e) What is the force on the negative plastic ball? (f) You discover that the source of the electric field at location A is a negatively charged particle. Which of the numbered locations in Figure 13.64 shows the location of this negatively charged particle, relative to location A?

Short answer

a) The arrow indicating electric field is h arrow

b) The electric field at point A is$\left(0.8\times {10}^4,-0.8\times {10}^4,0\right)\mathrm{N}/\mathrm{C}$ .

c) The magnitude of electric field vector is$1.13\times {10}^4\mathrm{N}/\mathrm{C}$ .

d) The arrow h indicates the direction of force so the direction of electric filed is given by the arrow opposite to it that is arrow d in figure 13.65.

e) The force exerted on the plastic ball is$\left(-4.8\times {10}^{-5},4.8\times {10}^{-5},0\right)\mathrm{N}$ .

f) The location of negatively charged particle is location 3.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The charge on the glass ball is,$\mathrm{q}=5\times {10}^{-9}\mathrm{C}$ .
• Force experienced by glass ball is,${\mathrm{F}}_{\mathrm{f}}=\left(4\times {10}^{-5},-4\times {10}^{-5},0\right)\mathrm{N}$
• Force experienced by plastic ball is,${\mathrm{q}}_{\mathrm{p}}=-6\times {10}^{-9}\mathrm{C}$ .

Concept/Significance of electric field

When a charge encounters a force in a certain region of space, it creates an electric field. If the electric field remains constant throughout time and space, it means that neither its direction nor amplitude changes.

(a) Determination of arrow in Figure 13.65 best indicates the direction of the electric field at location A

The electric field have the same direction to the direction of force on positive charge particle.so the arrow that indicates electric force will also show the direction of electric field.

Thus, the arrow indicating electric field is h arrow.

(b) Determination of the the electric field at location A

The electric field is given by,

$\overrightarrow{E}=\frac{\overrightarrow{F}}{q}$

Here, $\overrightarrow{F}$is the electric force exerted on particle and q is the charge on the glass ball.

Substitute all the values in the above,

role="math" localid="1656920153956" $$\begin{gathered} \overrightarrow{E}=\frac{\left(4\times {10}^{-5},-4\times {10}^{-5},0\right)\mathrm{N}}{5\times {10}^{-9}\mathrm{C}} \\ =\left(0.8\times {10}^4,-0.8\times {10}^4,0\right)\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the electric field at point A is $\left(0.8\times {10}^4,-0.8\times {10}^4,0\right)\mathrm{N}/\mathrm{C}$.

(c) Determination of the magnitude of this electric field

The magnitude of electric field is given by,

$\left|\overrightarrow{E}\right|=\sqrt{{\left(E_x\right)}^2+{\left(E_y\right)}^2+{\left(E_z\right)}^2}$

Here, $E_x$is the component of electric field in x-direction,$E_y$ is the component of electric field in y-direction,$E_z$ is the component of electric field in z-direction.

Substitute all the values in the above,

$$\begin{gathered} \left|\overrightarrow{E}\right|=\sqrt{\left(0.8\times {10}^4\mathrm{N}/\mathrm{C}\right)+{\left(0.8\times {10}^4\mathrm{N}/\mathrm{C}\right)}^2+0} \\ =1.13\times {10}^4\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the magnitude of electric field vector is$1.13\times {10}^4\mathrm{N}/\mathrm{C}$ .

(d) Determination of the arrow in Figure 13.65 best indicates the direction of the electric force on the negatively charged plastic ball when the glass ball is moved very far away and a tiny plastic ball with charge -6×10-9C is placed at location A

The electric file has the opposite direction to the force on the negative charge because it depends on the magnitude of the charge. So, the arrow opposite to the arrow indicating the direction of force will give the direction of electric field.

Thus, the arrow h indicates the direction of force so the direction of electric filed is given by the arrow opposite to it that is arrow d in figure 13.65.

(e) Determination of the force on the negative plastic ball

The force on the negative ball is given by,

$\overrightarrow{F}=\overrightarrow{E}{q}_p$

Here,$\overrightarrow{E}$ is the electric field and$q_p$ is the charge on plastic ball.

Substitute all the values in the above,

role="math" localid="1656921121510" $$\begin{gathered} \overrightarrow{\mathrm{F}}=\left(0.8\times {10}^4,-0.8\times {10}^4,0\right)\mathrm{N}/\mathrm{C}\left(-6\times {10}^{-9}\mathrm{C}\right) \\ =\left(-4.8\times {10}^{-5},4.8\times {10}^{-5},0\right)\mathrm{N} \end{gathered}$$

Thus,the force exerted on the plastic ball isrole="math" localid="1656921129920" $\left(-4.8\times {10}^{-5},4.8\times {10}^{-5},0\right)\mathrm{N}$

(e) Determination of the number of locations in Figure 13.64 shows the location of the negatively charged particle, relative to location A

The location of negative charge is described by the direction of electric field and electric force. In figure 13.64 the direction of electric force is towards location 1.

Thus, the location of negatively charged particle is location 3.